AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
. Prove that
Let 
be reals such that 
. Prove that
+2 w
be the side lengths of a triangle. Prove that![\[\frac{ab+bc+ca}{(a+b)^2} + \frac{ab+bc+ca}{(b+c)^2} + \frac{ab+bc+ca}{(c+a)^2} \leqslant \frac{85}{36}.\]](http://latex.artofproblemsolving.com/f/8/0/f80aed170f4f4bd78a2e881ce93b6b7ddfa03c22.png)
such that the reflection of each of these four points about lies on the circle passing through three remaining points.
, the following inequality always holds:![\[
\frac{ab}{b+1} + \frac{bc}{c+1} + \frac{ca}{a+1} \geq \frac{3abc}{1 + abc}
\]](http://latex.artofproblemsolving.com/d/5/8/d582a3269c92041834c02153ee8e946a274e52bd.png)
and 
be positive integers. is a divisor of 
and is a divisor of 
.
., find all possible positive integer values of![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
be a triangle with 
and 
.
inside triangle satisfies the following conditions:![\[
XA^2 + BC^2 \leq XC^2 + AB^2 \leq XB^2 + AC^2.
\]](http://latex.artofproblemsolving.com/0/3/1/031aa58d7d3dabe636dfb4245ef2456a2a3d5df6.png)
Find the maximum value of 
such that![\[
\frac{XA}{AC} \geq m.
\]](http://latex.artofproblemsolving.com/c/5/e/c5ed2892b224aff3b2f3e20b4c9424707ff88690.png)
be a cyclic quadrilateral an let 
be a point on the side 
The diagonals 
meets the segments 
at 
The line through parallel to 
mmets the extension of the side 
beyond 
at 
The line through 
parallel to 
meets the extension of the side beyond at 
Prove that the circumcircles of the triangles 
and 
are tangent .
be non-negative real numbers such that 
Prove that
board. On each cell of the main diagonal there is a file. At each step, we may select some files and move them to the left, on their respective rows, by the same distance. What is the minimum number of necessary moves in order to put all files on the first column? be the side lengths of a triangle. Prove that![\[\frac{a^3}{(a+b)^3}+\frac{b^3}{(b+c)^3}+\frac{c^3}{(c+a)^3} \leqslant \frac{9}{8}.\]](http://latex.artofproblemsolving.com/5/1/5/515cbd3cd69fee483312f90d4cb29d98ca761f41.png)
spheres centered at 
, with radius of 
and , which is remarked as 
and 
, respectively. Given 
points 
outside of (not including the surface of ).
as the number of sets 
such that the midpoint of 
lies entirely inside of .
I thought it was 2 n's
?
?
?
where 
divisible by 
prime numbers prime
is a prime number, show that 
is composite.
or 
.
,
,
which is impossible, so 
.
.
such that 
is a perfect cube.
is the only solution. It can easily be verified that it is indeed one; we will prove that it is the only one.
,
is odd, therefore 
,
This means that the order of 
is a number that divides 
,
and not 
which implies that 
,
;
is the only solution.
.
.
,
(
)
.
.
.
.
.
,
.
,
i
a
that satisfy 
such that 
.![\[
9 \cdot 2^m + 1 = q^5,
\]](http://latex.artofproblemsolving.com/e/2/e/e2e257fe01262176977bfcceee9df4c96d02c4c8.png)
so![\[
9 \cdot 2^m = (q - 1)(q^4 + q^3 + q^2 + q + 1).
\]](http://latex.artofproblemsolving.com/5/f/a/5fa978669eecac8802111f79ebb45f38ddb9041c.png)
Thus, 
is divisible by 
.
,
.![\[
2^m \cdot p^2 = (q - 1)(q^4 + q^3 + q^2 + q + 1).
\]](http://latex.artofproblemsolving.com/7/6/c/76c4aa4bb2069940af900eb4c1ff490c049e3647.png)
must be divisible by 
and it's odd, 
must be odd and there are only two possibilities:
,
,
and 
.
,![\[
2^m \cdot p = q - 1 \Rightarrow 2^m \cdot (q^4 + q^3 + q^2 + q + 1) = q - 1.
\]](http://latex.artofproblemsolving.com/e/5/2/e52d97b29dde88264beb4d2e1852da4d1d033ac9.png)
is a power of 3.
.
,
for 
.
,
,
for some 
.
and 
,
,
.
satisfies the following recurrence relation.
Observing this 
and 
,
iff 
,
iff 
implies 
,
.
of positive integers, such that ![\[a^3 + b^3 = p^n\mbox{.}\]](http://latex.artofproblemsolving.com/a/7/b/a7bbc61dc8020b99d31957c00e3ac85b139e8a62.png)
and 
where 
are nonnegative integers.
.
.
,
or 
.
by 
,
.
.
,
,
is the smallest 
and 
.
,
is also a root. If we don’t reach 
doing 
,
(
is the case 
)
in 
,
and similarly 
.
where 
is nonnegative integer.
for which there exist a positive integers 